Question

In the given parallelogram ABCD, O is mid-point of AB. CO and DO bisect $\angle BCD$ and $\angle ADC$ respectively. Find the measure of $\angle DOC$.

Answer 1

In parallelogram ABCD,
$\angle ADC+\angle BCD={180}^{\circ }$
Adjacent angles of a parallelogram is supplementary
$\frac{\angle ADC}{2}+\frac{\angle BCD}{2}={90}^{\circ }$
$\angle ODC+\angle DCO={90}^{\circ }...\left(i\right)$
Also, the sum of all the angles of a triangle is ${180}^{\circ }$
So, $\angle DOC+\angle ODC+\angle DCO={180}^{\circ }$
From (i)
$\angle DOC+{90}^{\circ }={180}^{\circ }$
$\angle DOC={180}^{\circ }-{90}^{\circ }$
$\angle DOC={90}^{\circ }$